Problem: Simplify; express your answer in exponential form. Assume $x\neq 0, k\neq 0$. $\dfrac{{(x^{-3})^{2}}}{{(x^{-1}k^{-2})^{-1}}}$
To start, try working on the numerator and the denominator independently. In the numerator, we have ${x^{-3}}$ to the exponent ${2}$ . Now ${-3 \times 2 = -6}$ , so ${(x^{-3})^{2} = x^{-6}}$ In the denominator, we can use the distributive property of exponents. ${(x^{-1}k^{-2})^{-1} = (x^{-1})^{-1}(k^{-2})^{-1}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(x^{-3})^{2}}}{{(x^{-1}k^{-2})^{-1}}} = \dfrac{{x^{-6}}}{{xk^{2}}}$ Break up the equation by variable and simplify. $\dfrac{{x^{-6}}}{{xk^{2}}} = \dfrac{{x^{-6}}}{{x}} \cdot \dfrac{{1}}{{k^{2}}} = x^{{-6} - {1}} \cdot k^{- {2}} = x^{-7}k^{-2}$.